Chapter 8, Problem 8.69EP

(a)

Interpretation Introduction

Interpretation:

The final concentration of the prepared solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.69EP

The final concentration of the prepared solution is 3.0 M.

Explanation of Solution

Given information:

Initial volume of solution = 30.0 mL

Initial concentration = 5.0 M

Final volume of solution = 50 mL (after 20.0 mL water is added)

Final concentration = ?

Apply dilution equation to calculate the molarity of the concentrated solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \\ {\text{C}}_{\text{d}}\text{=}\text{final}\text{concentration}\text{=}{\text{C}}_{\text{s}}\text{×}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{d}}}\\ \text{=}\text{5}\text{.0}\text{M}\text{×}\frac{\text{30}\text{mL}}{\text{50}\text{mL}}\text{=}\text{3}\text{.0}\text{M}\end{array}$

Therefore, the final concentration of the prepared solution is 3.0 M.

(b)

Interpretation Introduction

Interpretation:

The final concentration of the prepared solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.69EP

The final concentration of the prepared solution is 3.0 M.

Explanation of Solution

Given information:

Initial volume of solution = 30.0 mL

Initial concentration = 5.0 M

Final volume of solution = 50 mL (after 20.0 mL water is added)

Final concentration = ?

Apply dilution equation to calculate the molarity of the concentrated solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \\ {\text{C}}_{\text{d}}\text{=}\text{final}\text{concentration}\text{=}{\text{C}}_{\text{s}}\text{×}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{d}}}\\ \text{=}\text{5}\text{.0}\text{M}\text{×}\frac{\text{30}\text{mL}}{\text{50}\text{mL}}\text{=}\text{3}\text{.0}\text{M}\end{array}$

Therefore, the final concentration of the prepared solution is 3.0 M.

(c)

Interpretation Introduction

Interpretation:

The final concentration of the prepared solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.69EP

The final concentration of the prepared solution is 4.5 M.

Explanation of Solution

Given information:

Initial volume of solution = 30.0 mL

Initial concentration = 7.5 M

Final volume of solution = 50 mL (after 20.0 mL water is added)

Final concentration = ?

Apply dilution equation to calculate the molarity of the concentrated solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \\ {\text{C}}_{\text{d}}\text{=}\text{final}\text{concentration}\text{=}{\text{C}}_{\text{s}}\text{×}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{d}}}\\ \text{=}\text{7}\text{.5}\text{M}\text{×}\frac{\text{30}\text{mL}}{\text{50}\text{mL}}\text{=}\text{4}\text{.5}\text{M}\end{array}$

Therefore, the final concentration of the prepared solution is 4.5 M.

(d)

Interpretation Introduction

Interpretation:

The final concentration of the prepared solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.69EP

The final concentration of the prepared solution is 1.5 M.

Explanation of Solution

Given information:

Initial volume of solution = 60.0 mL

Initial concentration = 2.0 M

Final volume of solution = 80 mL (after 20.0 mL water is added)

Final concentration = ?

Apply dilution equation to calculate the molarity of the concentrated solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \\ {\text{C}}_{\text{d}}\text{=}\text{final}\text{concentration}\text{=}{\text{C}}_{\text{s}}\text{×}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{d}}}\\ \text{=}\text{2}\text{.0}\text{M}\text{×}\frac{\text{60}\text{mL}}{\text{80}\text{mL}}\text{=}\text{1}\text{.5}\text{M}\end{array}$

Therefore, the final concentration of the prepared solution is 1.5 M.